Problem F: STL——集合运算

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Problem F: STL——集合运算

Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 3767  Solved: 1927 [ Submit][ Status][ Web Board]

Description

集合的运算就是用给定的集合去指定新的集合。设A和B是集合，则它们的并差交补集分别定义如下： A∪B={x|x∈A∨x∈B} A∩B={x|x∈A∧x∈B} A-B={x|x∈A∧x不属于 B} SA ={x|x∈(A∪B)∧x 不属于A} SB ={x|x∈(A∪B)∧x 不属于B}

Input

第一行输入一个正整数T，表示总共有T组测试数据。（T<=200） 然后下面有2T行，每一行都有n+1个数字，其中第一个数字是n(0<=n<=100)，表示该行后面还有n个数字输入。

Output

对于每组测试数据，首先输出测试数据序号，”Case #.NO”， 接下来输出共7行，每行都是一个集合， 前2行分别输出集合A、B，接下5行来分别输出集合A、B的并(A u B)、交(A n B)、差(A – B)、补。 集合中的元素用“{}”扩起来，且元素之间用“， ”隔开。

Sample Input

14 1 2 3 10

Sample Output

Case# 1:A = {1, 2, 3}B = {}A u B = {1, 2, 3}A n B = {}A - B = {1, 2, 3}SA = {}SB = {1, 2, 3}

HINT

如果你会用百度搜一下关键字“stl set”，这个题目我相信你会很快很轻松的做出来。加油哦！

Append Code

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#include <iostream> #include <algorithm> #include <set> using namespace std; int main() { int T, j, n, a[110]; cin >> T; for(int i = 1; i <= T; i++) { set <int> A; set <int> B; set <int> C; A.clear(); B.clear(); C.clear(); set <int>:: iterator set_iter; cin >> n; for( j = 0; j < n; ++j) cin >> a[j]; A.insert(a, a+n); cin >> n; for(int j = 0; j < n; j++) cin >> a[j]; B.insert(a, a + n); cout << "Case# " << i << ":" << endl; cout << "A = {"; for(set_iter = A.begin(); set_iter != A.end() ; ++set_iter) { if(set_iter != A.begin()) cout << ", "; cout << *set_iter; } cout << "}" <<endl; cout << "B = {"; for(set_iter = B.begin(); set_iter != B.end() ; ++set_iter) { if(set_iter != B.begin()) cout << ", "; cout << *set_iter; } cout << "}" <<endl; cout << "A u B = {"; set_union(A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int > >(C,C.begin())); for(set_iter = C.begin(); set_iter != C.end() ; ++set_iter) { if(set_iter != C.begin()) cout << ", "; cout << *set_iter; } cout << "}" <<endl; C.clear(); set_intersection( A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin())); cout << "A n B = {"; for(set_iter = C.begin(); set_iter != C.end() ; ++set_iter) { if(set_iter != C.begin()) cout << ", "; cout << *set_iter; } cout << "}" <<endl; C.clear(); cout << "A - B = {"; set_difference(A.begin(), A.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin())); for(set_iter = C.begin(); set_iter != C.end(); set_iter++ ) { if(set_iter != C.begin()) cout << ", "; cout << *set_iter; } cout << "}" << endl; C.clear(); set<int> C2; cout << "SA = {"; set_union(A.begin(),A.end(),B.begin(),B.end(),insert_iterator<set<int> >(C2,C2.begin())); set_difference(C2.begin(), C2.end(), A.begin(), A.end(), insert_iterator< set<int> >(C, C.begin())); for(set_iter = C.begin(); set_iter != C.end(); set_iter++ ) { if(set_iter != C.begin()) cout << ", "; cout << *set_iter; } cout << "}" << endl; C.clear(); cout << "SB = {"; set_difference(C2.begin(), C2.end(), B.begin(), B.end(), insert_iterator< set<int> >(C, C.begin())); for(set_iter = C.begin(); set_iter != C.end(); set_iter++ ) { if(set_iter != C.begin()) cout << ", "; cout << *set_iter; } cout << "}" << endl; A.clear(); B.clear(); C.clear(); C2.clear(); } }